Objective: What is the speed of a Constant Motion Vehicle (CMV)? Hypothesis
The CMV is going 3 cm per second.
The position time graph tells us the relationship between the position of the car and how long it took to get there.
You are required to measure very precisely.
Data Table and Position Time Graph- Speed of Yellow CMV Analysis The slope is the average speed. The meaning of the equation of the line is v=d/t. R2 tells if the line is a good fit into your data. Discussion questions 1. Why is the slope of the position-time graph equivalent to average velocity? The slope of the position-time graph is equivalent to the average velocity because slope is centimeters/seconds and the average velocity was the number of centimeters over 1. 2. Why is it average velocity and not instantaneous velocity? What assumptions are we making? It is the average velocity because its an average of all the positions and seconds, if it was instantaneous than the velocity would be the same for each position, but as we see they are not all equal. 3. Why was it okay to set the y-intercept equal to zero? It is okay to set the y-intercept equal to zero because we started the experiment at 0 where the position and seconds both were equal to zero. 4. What is the meaning of the R2 value? R2 tells if the line is a good fit into your data. If it is above 95% then your data is on the path it should be. 5. If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours? The results of the second CMV would be similar to the line our CMV created but it would lie beneath our line because the position would be lower but the time would be the same.
Conclusion Through the experiment we found the results to be that the yellow slow CMV moves at 10.538 cm per second. Both of our hypotheses were much lower than the actual results, we believed that the CMV traveled at a much slower rate of 2 or 3 cm per second. We both saw the relationship between the positions of the CMV and the time it took to get there, that the graph created as well as the precision needed in order to measure. The main sources that contributed to inaccuracies were that the CMV’s were not identical due to their differing battery strengths, the elevated rulers made it difficult to measure, and the CMV did not drive in a straight line. In order to minimize these issues, we would put new, identical batteries in each CMV, use a flat or translucent ruler, and test the CMV on a flatter more level surface.
9/8/11 Lesson 1 (A,B,C,D) Summarizing Method 2a 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. In one of the documents they discussed distance and displacement. From our class discussion I already understood the difference and definition of each concept. I knew that distance is the total amount traveled and displacement is how much one traveled from where they start to where they end, it does not take into count the amount traveled in between. I was also aware of the formula used to calculate average speed before reading these documents. 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. From reading I realized that a main difference between vectors and scalars and speed and velocity. Scalars do not require a direction while vectors do, and velocity is direction aware and speed is not. 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understood all the concepts discussed in the documents. 4. What (specifically) did you read that was not gone over during class today? Displacement, distance, speed, and velocity were thoroughly gone over in class, where as the concepts of vectors and scalars were not really discussed.
Class Notes-9/9/11
Average speed--total distance and total time
Constant speed--same speed all the time
Instantaneous speed--speed your going at that moment
Equation--V=change in distance/change in time (m/s)
Signs are arbitrary (random)--someone made it up and we agree with it Arrow points down--negative sign Arrow points up--positive sign
9/9/11 Lesson 2 (A,B,C) Summarizing Method 2a 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. In our lab we learned a lot about Ticker Tape Diagrams. I understood that the distance between the lines depicted how fast or slow the object was moving. I also knew that the "oil stain" displayed on the diagram symbolized the stopping of the object, because that occurred with my CMV. On Friday we also already learned about vector diagrams and the way they looked when displaying constant speed and accelerating speed. 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. I wasn't shaky about many things we learned in class, so the reading just reassured my prior knowledge. 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understood all parts of the reading because I felt that I knew most of the information from class. 4. What (specifically) did you read that was not gone over during class today? All the information was gone over during class.
9/12/11 Partner: Rachel Knapel Activity: Graphical Representations of Equilibrium
How can you tell that there is no motion on a…
position vs. time graph—The line is straight
velocity vs. time graph—The line is on zero
acceleration vs. time graph—The line is on zero
How can you tell that your motion is steady on a…
position vs. time graph—The slope is constant
velocity vs. time graph—The line is at zero
acceleration vs. time graph—The line is at zero and is not increasing nor decreasing
How can you tell that your motion is fast vs. slow on a…
position vs. time graph—Slope on fast line is steeper
velocity vs. time graph—The line is shorter for the same distance
acceleration vs. time graph—The line is shorter for the same distance
How can you tell that you changed direction on a…
position vs. time graph—Opposite slopes and starting position is closer because it is a smaller distance away from the Motion Detector
velocity vs. time graph—Walking away is above the x axis and walking towards the Motion Detector is below the x axis
acceleration vs. time graph—You were not able to tell the difference
What are the advantages of representing motion using a…
position vs. time graph—It is easer to see the differences in speed and direction
velocity vs. time graph—You can see not only the speed but also the distance.
acceleration vs. time graph—You can see the difference in speeds between the two runs.
What are the disadvantages of representing motion using a…
position vs. time graph—If you sway or don’t stay on the same path then the graph isn’t linear.
velocity vs. time graph—It is difficult to read
acceleration vs. time graph—If your speed changes at all there are spikes in the graph
Define the following:
No motion—when your acceleration and velocity are zero and your position is constant
Constant speed—you do not speed up or slow down, and your position has the and unchanged slope
This graph shows when Rachel was at rest.
This graph shows constant speed going towards and away the motion detector. Run 1 represents when i was walking away from the motion detector, while run 2 depicts when i was walking toward it. These graphs show constant motion slow vs. fast. Run 1 was the slow one, while run 2 was the fast one.
Kinematics - The Big 5 Notes
acceleration- rate that velocity is changing (a) m/s ^ 2
v = d/t ONLY for average or constant speeds
v = (vi + vf) / 2 ONLY for average speed
a = (vf-vi) / t
1/2(vi + vf) = change in d / change in t
vi = initial velocity
vf = final velocity
vf = vi + a(t) - has no change in d
change in d = 1/2(vi + vf)t - has no a
change in d = vi(t) + 1/2(a)(t^2) - has no vf
vf^2 = vi^2 + 2(a)(change in d) - has no t
9/13/11
Class Notes
At Rest and Constant Speed Graphs
What does a position-time graph for increasing speeds look like?
What information can be found from the graph?
Hypotheses: 1) The slope gets steeper as the cart approached the bottom of the ramp. 2) The graph shows the relationship between the cart and its speed at a certain point on the ramp.
Analysis: a) Interpret the equation of the line (slope, y-intercept) and the R2 value.--On graph b) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.) This picture displays our calculations for instantaneous speed of the halfway point and last point of the line.
c) Find the average speed for the entire trip.--On graph This is our data and our graph.
This video shows the car all during our experiment. It has the ticker tape taped to the back so as it rides along the ramp a dot is made every .1 of a second. THe car starts at rest then gets faster and faster as it goes down the ramp until it is eventually stopped by an outside force.
Discussion Questions: 1. What would your graph look like if the incline had been steeper? It would look more like the linear line of best fit and less like the polynomial line of best fit and it would have a steeper slope. 2. What would your graph look like if the cart had been decreasing up the incline? It would be steeper at the bottom and then curve as it approached the top since it will slow down as it goes up the incline. 3.Compare the instantaneous speed at the halfway point with the average speed of the entire trip. The average speed and the instantaneous speed at the halfway point were only 7.59 cm/s away from one another. The average speed and the instantaneous speed both represent the middle. The instantaneous being the median and the average speed being the mean 4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? Because instantaneous speed is what speed you are at that second so when the tangent intersects with a point on the graph, it is the slope for that point only. So therefore the instantaneous speed equals the slope of the tangent line. 5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
Conclusion: As a result, the further along the car was on the ramp the higher the speed was. As the car approached the bottom of the ramp, the speed gradually increased. The ticker tape dots got farther apart because of this, and this proved out hypothesis. We were correct in thinking that the graph shows a steeper slope as the cart approached the bottom of the ramp and that the graph would show the relationship between the carts position and speed. The ticker tape sometimes does not go through the spark timer in a straight line, causing faults in the data. Also, when releasing the cart, we could have added force, which would increase the speed. We would make sure the tape was lined up straighter before beginning if we could redo the lab and we would try to release the cart without adding any extra force.
9/14/11 Lesson 1 (E) Summarizing Method 2a 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. I fully understood the acceleration formulas discussed in the reading, because we learned them in class. As well, I understood the definition of acceleration, that it requires a change in velocity. 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? I was a little confused between the difference between constant velocity and constant acceleration, but the reading definitely helped me understand the separation of the two. 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. Why do you square the time of the free falling object? 4. What (specifically) did you read that was not gone over during class today? The acceleration of a free falling object
9/15/11
Class Notes
Increasing and Decreasing Speed Graphs
9/15/11 Lesson 3 (A,B,C) Summarizing Method 2a 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. I understood the relationship between velocity and slope on a position time graph and how the velocity and slope are directly related in reading the graph and determining the motion and direction of the object. I also knew what the graphs displaying acceleration as opposed to constant speed looked like, as well as the negative graphs of both. 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? The reading helped clarify how to determine if an object were accelerating on a position time graph as well as seeing if it were traveling in a positive or negative direction.The reading helped to clarify that the slope and velocity are equal, so when you find the slope with the formula for slope, you also find velocity. 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understood everything in the reading 4. What (specifically) did you read that was not gone over during class today? Everything discussed was gone over during class
9/15/11 Lesson 4 (A,B,C,D,E) Summarizing Method 2a 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. In class we had already seen the velocity time graphs that compared objects moving in a positive or negative direction. I already knew that if an object were moving in a negative direction they would be located in the negative region and if they were moving in a positive direction, they would be in the positive region. I also fully understood the direct relationship between acceleration and slope, and how they are equal. 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? I was a little confused about how to determine if an object were speeding up or slowing down, but the reading explained that if a line is moving away from the x axis then it is speeding up and if it is moving towards it then it is speeding up. 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understood all parts of the reading 4. What (specifically) did you read that was not gone over during class today? All parts were discussed in class, except finding area on a v-t graph.
Lab: A Crash Course In Velocity (Part II)
Date: 9/21/11 Partners: Rachel Knapel, Lindsay Marella, Maggie Leffler Objectives: Both algebraically and graphically, solve the following 2 problems. Then set up the situation and run trials to confirm your calculations.
Calculations for part A:
According to these calculations, the blue, faster car should travel 135.416 cm or about 1.4 m before catching up and becoming parallel with the yellow, slower car. Below is a video demonstrating this situation.
The blue car placed one meter behind the yellow car and is catching up to it at certain point. We then marked the spot where they met and saw that our calculations from before were close to the experiments results.
Calculations for part B:
Our calculations show that when the two cars meet, the blue should travel 475.375 cm and the yellow car should travel 124.626 cm. Below is a video of the actual situation of the two cars meeting up and crashing.
Here, the two cars crash at a certain point when separated 600 cm. We then marked the place where they crashed, and saw that our results were approximately correct.
Discussion Questions: 1. Where would the cars meet if their speeds were exactly equal?
If one car started in front of the other, the two cars would never meet. If they were facing each other and going the same speed than they would meet in the middle. 2. Sketch position-time graphs to represent the catching up and crashing situations. Show the points where they are at the same place at the same time. 3. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time?
Our CMV results were inaccurate, our percents of error were extremely high, however our percent difference was a lot closer to 0%. Our calculations show that when the two cars meet, the blue should travel 475.375 cm and the yellow car should travel 124.626 cm. Unfortunately, our results were not in agreement with our findings which caused a percent error which was very high. According to our calculations, the blue, faster car should have travel 135.416 cm or about 1.4 m before catching up and becoming parallel with the yellow, slower car. The average speed in our experiments was a much higher value of 256 cm. This also caused an elevated percent error. One source of error was probably that two people were turning on the CMVs; both of them could not possibly turn them on at the same time. This would mean that the CMV turned on faster would have had more time to gain more distance than the other one. This could be fixed by using a machine to turn on both at the same time, but this is not practical. Another source of error may have been from the curvature of the path of the CMV. Some of the CMVs curved to the left or right, and did not go directly straight; this would make it have to travel more distance in order to reach the point of collision or catch up. This could be fixed by using a track to guide the CMVs on a straight track. One other source of error may have been the fact that the batteries continued to fall out. This could be fixed by putting a back on the batteries, which was not present during our trials.
Practice Problem 9/22/11
Egg Drop Project
This is our egg drop project. It is a cone made out of paper with straws at the bottom and top to secure the egg from wiggling or moving.
Our results when we dropped our contraption from the roof was that the egg survived, was intact, and did not break.
Analysis: When we calculated the acceleration we got a very high number of 14.05m/s which is impossible. It should be close to or around 9.8m/s, but it was not. There may be flaws in our calculations. Also, the fact that their were only two times to find the average time may have ruined our calculations. Or maybe one person took the time down wrong. Molly may have released the cone and then the person timing may have started their clock. There are many faults that could have happened to lead to this miscalculation. In the experiment i would not have done much different since the fact that it worked. I might have used less paper to try and make it weigh less and also tighten up the cone so that i wouldn't need the straws to keep it in place.
10/3/11
Class Notes
Freefall-an object moving under the influence of gravity ONLY
Practice Problem
10/3/11
Lesson 5 (A,B,C,D,E)
Introduction to Free Fall
Topic sentence: Gravity is the only only influence on free falling objects, which accelerate downward at a rate or 9.8 m/s/s.
A free falling object is an object that is falling under the sole influence of gravity. They do not encounter air resistance. All free-falling objects accelerate downwards at a rate of 9.8 m/s/s. Because of this, a ticker tape trace diagram would depict an acceleration and if an object travels downward and speeds up, then its acceleration is downward.
The Acceleration of Gravity
Topic sentence: The acceleration of gravity is shown as g and is numerically equal to 9.8 m/s/s.
9.8 m/s/s is known as the acceleration of gravity and is symbolically shown as g. We will occasionally use the approximated value of 10 m/s/s in in order to reduce the complexity of the many mathematical tasks that we will perform with this number.
Representing Free Fall by Graphs
Topic sentence: Free falling objects have a negative acceleration.
One mean of describing the motion of objects is through the use of graphs -position versus time and velocity vs. time graphs.
A curved line on a position versus time graph signifies an accelerated motion. A further look at the position-time graph reveals that the object starts slow and finishes fast. Finally, the negative slope of the line indicates a negative or downward velocity.
A diagonal line on a velocity versus time graph signifies an accelerated motion. The object is moving in the negative direction and speeding up with a negative acceleration.
How Fast? and How Far?
Topic sentence: The velocity of a free falling object is dependent on the time it fell and the time of the fall.
The velocity of a free-falling object is changing by 9.8 m/s every second. The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is----vf = g * t g is 9.8 m/s/s.
The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of t seconds is given by---d = 0.5 * g * t2
The Big Misconception
Topic sentence: A more massif object does not accelerate at a greater rate than a less massive object.
A more massive object does not accelerate at a greater rate than a less massive object. More massive objects will only fall faster if there is an appreciable amount of air resistance present. The actual explanation of why all objects accelerate at the same rate involves the concepts of force and mass. The greater force on more massive objects is offset by the inverse influence of greater mass, but all objects free fall at the same rate of acceleration, regardless of their mass.
10/4/11
Free Fall Lab
100 g weight
The equation of the velocity time graph of y=mx+b is the same as the formula v(f)=v(i)+at. m or the slope has the same value as a, or acceleration and b or the y intercept is equal to the initial velocity. Our slope or acceleration is 775.37 and inital velocity or y intercept is 52.763.
In the position time graph, the equation is y=Ax^2+Bx, is also equal to the equation d=v(i)t+.5at^2. In this equation A has the same value as half the acceleration and B has the same value as the initial velocity. In our equation, half our velocity or A value is 387.93 and initial velocity or B value is 52.006. The initial velocity should be zero but is not because the timer and the drop of the object might not have been at the same time, the object could have been in motion when the spark timer made the first dot.We used a polynomial shape trend line to also show that our results fit the trend line almost perfectly. Additionally, our r squared value is .99 which shows that our results were precise and correct.
Error is due to the friction that occurred with the spark timer.
Discussion Questions 1. Does the shape of your v-t graph agree with the expected graph? Why or why not?
The shape of our v-t graph fully agrees with the expected graph because they are both linear due to constant speed. Unfortunately, because the object is being dropped, the line should be moved negatively away from the origin. Because the negatives were left out of our data our graph turned out the way it did. 2. Does the shape of your x-t graph agree with the expected graph? Why or why not?
Yes, our x-t graph agrees with the expected graph, because the object accelerated downward and this is displayed in the curved graph. 3. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
Our final result was that the acceleration was 775.37 cm/s/s and the class average was 805.87 cm/s/s. After finding the percent difference, we found that we have a 3.78% difference. Our results were below the 5% difference range, which implies that they were very close to equal the class average. 4. Did the object accelerate uniformly? How do you know?
Yes, this is shown in our v-t graph because it is seen that it has a constant acceleration. This is because of the linear display that comes from our data. Although it did not accelerate at exactly 981 cm/s/s, it accelerated at a constant rate which was pretty close to this. This shows that it did accelerate uniformly. 5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
Additional force from the person dropping the weight could cause a higher acceleration due to gravity. However, in our case the lower acceleration of gravity was caused by the friction of the ticker tape with the spark timer as it was threaded through.
Conclusion:
Our final results were that our weight fell with an acceleration of 775.37 cm/s/s. However, the desired or theoretical result was 981 cm/s/s. We then found out that our percent error was 20.96% which implies accuracy, but unfortunately not precision. We also found that the percent difference from the class average was 3.78%, which is very close the class average proving that our results were almost accurate. Our hypothesis was also correct, however, we kept our acceleration due to gravity in meters rather than centimeters and we said that the result should be negative. This conclusion is correct because the object is falling down, but for the experiment we disregarded the negative. One main source of error in our experiment was the friction from the ticker tape going through the spark timer. This would cause the acceleration to be lower. To fix this for future trials, you could factor in the lower acceleration do to friction. There was also added error because human's are not machines, which causes an additional area of human error. Because I had to drop the mass off the balcony I could have been a source of error because I could have slightly pushed the mass downwards, which would change the initial velocity. Also, Rachel or I could have measured the distance between the dots on the ticker tape incorrectly. To fix this error we could have used a different form of measurement that is more precise.
Table of Contents
Constant Speed
Lab: A Crash Course in Velocity (Part 1)
Partner: Rachel KnapelDate: 9/6/11
Objective: What is the speed of a Constant Motion Vehicle (CMV)?
Hypothesis
- The CMV is going 3 cm per second.
- The position time graph tells us the relationship between the position of the car and how long it took to get there.
- You are required to measure very precisely.
Data Table and Position Time Graph- Speed of Yellow CMVAnalysis
The slope is the average speed. The meaning of the equation of the line is v=d/t.
R2 tells if the line is a good fit into your data.
Discussion questions
1. Why is the slope of the position-time graph equivalent to average velocity?
The slope of the position-time graph is equivalent to the average velocity because slope is centimeters/seconds and the average velocity was the number of centimeters over 1.
2. Why is it average velocity and not instantaneous velocity? What assumptions are we making?
It is the average velocity because its an average of all the positions and seconds, if it was instantaneous than the velocity would be the same for each position, but as we see they are not all equal.
3. Why was it okay to set the y-intercept equal to zero?
It is okay to set the y-intercept equal to zero because we started the experiment at 0 where the position and seconds both were equal to zero.
4. What is the meaning of the R2 value?
R2 tells if the line is a good fit into your data. If it is above 95% then your data is on the path it should be.
5. If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours?
The results of the second CMV would be similar to the line our CMV created but it would lie beneath our line because the position would be lower but the time would be the same.
Conclusion
Through the experiment we found the results to be that the yellow slow CMV moves at 10.538 cm per second. Both of our hypotheses were much lower than the actual results, we believed that the CMV traveled at a much slower rate of 2 or 3 cm per second. We both saw the relationship between the positions of the CMV and the time it took to get there, that the graph created as well as the precision needed in order to measure. The main sources that contributed to inaccuracies were that the CMV’s were not identical due to their differing battery strengths, the elevated rulers made it difficult to measure, and the CMV did not drive in a straight line. In order to minimize these issues, we would put new, identical batteries in each CMV, use a flat or translucent ruler, and test the CMV on a flatter more level surface.
9/8/11
Lesson 1 (A,B,C,D)
Summarizing Method 2a
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
In one of the documents they discussed distance and displacement. From our class discussion I already understood the difference and definition of each concept. I knew that distance is the total amount traveled and displacement is how much one traveled from where they start to where they end, it does not take into count the amount traveled in between. I was also aware of the formula used to calculate average speed before reading these documents.
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
From reading I realized that a main difference between vectors and scalars and speed and velocity. Scalars do not require a direction while vectors do, and velocity is direction aware and speed is not.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
I understood all the concepts discussed in the documents.
4. What (specifically) did you read that was not gone over during class today?
Displacement, distance, speed, and velocity were thoroughly gone over in class, where as the concepts of vectors and scalars were not really discussed.
Class Notes-9/9/11
- Average speed--total distance and total time
- Constant speed--same speed all the time
- Instantaneous speed--speed your going at that moment
Equation--V=change in distance/change in time (m/s)Signs are arbitrary (random)--someone made it up and we agree with it
Arrow points down--negative sign
Arrow points up--positive sign
9/9/11
Lesson 2 (A,B,C)
Summarizing Method 2a
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
In our lab we learned a lot about Ticker Tape Diagrams. I understood that the distance between the lines depicted how fast or slow the object was moving. I also knew that the "oil stain" displayed on the diagram symbolized the stopping of the object, because that occurred with my CMV. On Friday we also already learned about vector diagrams and the way they looked when displaying constant speed and accelerating speed.
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding.
I wasn't shaky about many things we learned in class, so the reading just reassured my prior knowledge.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
I understood all parts of the reading because I felt that I knew most of the information from class.
4. What (specifically) did you read that was not gone over during class today?
All the information was gone over during class.
9/12/11
Partner: Rachel Knapel
Activity: Graphical Representations of Equilibrium
This graph shows when Rachel was at rest.
This graph shows constant speed going towards and away the motion detector. Run 1 represents when i was walking away from the motion detector, while run 2 depicts when i was walking toward it.
These graphs show constant motion slow vs. fast. Run 1 was the slow one, while run 2 was the fast one.
Kinematics - The Big 5 Notes
acceleration- rate that velocity is changing (a) m/s ^ 2
a = (vf-vi) / t
1/2(vi + vf) = change in d / change in t
vi = initial velocity
vf = final velocity
vf = vi + a(t) - has no change in d
change in d = 1/2(vi + vf)t - has no a
change in d = vi(t) + 1/2(a)(t^2) - has no vf
vf^2 = vi^2 + 2(a)(change in d) - has no t
9/13/11
Class Notes
At Rest and Constant Speed Graphs
Lab: Acceleration Graphs
9/13/11
Partner: Rachel Knapel
Objectives:
Hypotheses:
1) The slope gets steeper as the cart approached the bottom of the ramp.
2) The graph shows the relationship between the cart and its speed at a certain point on the ramp.
Available Materials:
Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape
Analysis:
a) Interpret the equation of the line (slope, y-intercept) and the R2 value.--On graph
b) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.)
This picture displays our calculations for instantaneous speed of the halfway point and last point of the line.
c) Find the average speed for the entire trip.--On graph
This is our data and our graph.
This video shows the car all during our experiment. It has the ticker tape taped to the back so as it rides along the ramp a dot is made every .1 of a second. THe car starts at rest then gets faster and faster as it goes down the ramp until it is eventually stopped by an outside force.
Discussion Questions:
1. What would your graph look like if the incline had been steeper?
It would look more like the linear line of best fit and less like the polynomial line of best fit and it would have a steeper slope.
2. What would your graph look like if the cart had been decreasing up the incline?
It would be steeper at the bottom and then curve as it approached the top since it will slow down as it goes up the incline.
3.Compare the instantaneous speed at the halfway point with the average speed of the entire trip.
The average speed and the instantaneous speed at the halfway point were only 7.59 cm/s away from one another. The average speed and the instantaneous speed both represent the middle. The instantaneous being the median and the average speed being the mean
4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense?
Because instantaneous speed is what speed you are at that second so when the tangent intersects with a point on the graph, it is the slope for that point only. So therefore the instantaneous speed equals the slope of the tangent line.
5. Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
Conclusion:
As a result, the further along the car was on the ramp the higher the speed was. As the car approached the bottom of the ramp, the speed gradually increased. The ticker tape dots got farther apart because of this, and this proved out hypothesis. We were correct in thinking that the graph shows a steeper slope as the cart approached the bottom of the ramp and that the graph would show the relationship between the carts position and speed. The ticker tape sometimes does not go through the spark timer in a straight line, causing faults in the data. Also, when releasing the cart, we could have added force, which would increase the speed. We would make sure the tape was lined up straighter before beginning if we could redo the lab and we would try to release the cart without adding any extra force.
9/14/11
Lesson 1 (E)
Summarizing Method 2a
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
I fully understood the acceleration formulas discussed in the reading, because we learned them in class. As well, I understood the definition of acceleration, that it requires a change in velocity.
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify?
I was a little confused between the difference between constant velocity and constant acceleration, but the reading definitely helped me understand the separation of the two.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
Why do you square the time of the free falling object?
4. What (specifically) did you read that was not gone over during class today?
The acceleration of a free falling object
9/15/11
Class Notes
Increasing and Decreasing Speed Graphs
9/15/11
Lesson 3 (A,B,C)
Summarizing Method 2a
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
I understood the relationship between velocity and slope on a position time graph and how the velocity and slope are directly related in reading the graph and determining the motion and direction of the object. I also knew what the graphs displaying acceleration as opposed to constant speed looked like, as well as the negative graphs of both.
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify?
The reading helped clarify how to determine if an object were accelerating on a position time graph as well as seeing if it were traveling in a positive or negative direction.The reading helped to clarify that the slope and velocity are equal, so when you find the slope with the formula for slope, you also find velocity.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
I understood everything in the reading
4. What (specifically) did you read that was not gone over during class today?
Everything discussed was gone over during class
9/15/11
Lesson 4 (A,B,C,D,E)
Summarizing Method 2a
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.
In class we had already seen the velocity time graphs that compared objects moving in a positive or negative direction. I already knew that if an object were moving in a negative direction they would be located in the negative region and if they were moving in a positive direction, they would be in the positive region. I also fully understood the direct relationship between acceleration and slope, and how they are equal.
2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify?
I was a little confused about how to determine if an object were speeding up or slowing down, but the reading explained that if a line is moving away from the x axis then it is speeding up and if it is moving towards it then it is speeding up.
3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question.
I understood all parts of the reading
4. What (specifically) did you read that was not gone over during class today?
All parts were discussed in class, except finding area on a v-t graph.
Lab: A Crash Course In Velocity (Part II)
Date: 9/21/11
Partners: Rachel Knapel, Lindsay Marella, Maggie Leffler
Objectives: Both algebraically and graphically, solve the following 2 problems. Then set up the situation and run trials to confirm your calculations.
Calculations for part A:
According to these calculations, the blue, faster car should travel 135.416 cm or about 1.4 m before catching up and becoming parallel with the yellow, slower car. Below is a video demonstrating this situation.
The blue car placed one meter behind the yellow car and is catching up to it at certain point. We then marked the spot where they met and saw that our calculations from before were close to the experiments results.
Calculations for part B:
Our calculations show that when the two cars meet, the blue should travel 475.375 cm and the yellow car should travel 124.626 cm. Below is a video of the actual situation of the two cars meeting up and crashing.
Here, the two cars crash at a certain point when separated 600 cm. We then marked the place where they crashed, and saw that our results were approximately correct.
Discussion Questions:
1. Where would the cars meet if their speeds were exactly equal?
If one car started in front of the other, the two cars would never meet. If they were facing each other and going the same speed than they would meet in the middle.
2. Sketch position-time graphs to represent the catching up and crashing situations. Show the points where they are at the same place at the same time.
3. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time?
Our CMV results were inaccurate, our percents of error were extremely high, however our percent difference was a lot closer to 0%. Our calculations show that when the two cars meet, the blue should travel 475.375 cm and the yellow car should travel 124.626 cm. Unfortunately, our results were not in agreement with our findings which caused a percent error which was very high. According to our calculations, the blue, faster car should have travel 135.416 cm or about 1.4 m before catching up and becoming parallel with the yellow, slower car. The average speed in our experiments was a much higher value of 256 cm. This also caused an elevated percent error. One source of error was probably that two people were turning on the CMVs; both of them could not possibly turn them on at the same time. This would mean that the CMV turned on faster would have had more time to gain more distance than the other one. This could be fixed by using a machine to turn on both at the same time, but this is not practical. Another source of error may have been from the curvature of the path of the CMV. Some of the CMVs curved to the left or right, and did not go directly straight; this would make it have to travel more distance in order to reach the point of collision or catch up. This could be fixed by using a track to guide the CMVs on a straight track. One other source of error may have been the fact that the batteries continued to fall out. This could be fixed by putting a back on the batteries, which was not present during our trials.
Practice Problem
9/22/11
Egg Drop Project
This is our egg drop project. It is a cone made out of paper with straws at the bottom and top to secure the egg from wiggling or moving.
Our results when we dropped our contraption from the roof was that the egg survived, was intact, and did not break.
Analysis: When we calculated the acceleration we got a very high number of 14.05m/s which is impossible. It should be close to or around 9.8m/s, but it was not. There may be flaws in our calculations. Also, the fact that their were only two times to find the average time may have ruined our calculations. Or maybe one person took the time down wrong. Molly may have released the cone and then the person timing may have started their clock. There are many faults that could have happened to lead to this miscalculation. In the experiment i would not have done much different since the fact that it worked. I might have used less paper to try and make it weigh less and also tighten up the cone so that i wouldn't need the straws to keep it in place.
10/3/11
Class Notes
Freefall-an object moving under the influence of gravity ONLY
Practice Problem
10/3/11
Lesson 5 (A,B,C,D,E)
Introduction to Free Fall
Topic sentence: Gravity is the only only influence on free falling objects, which accelerate downward at a rate or 9.8 m/s/s.A free falling object is an object that is falling under the sole influence of gravity. They do not encounter air resistance. All free-falling objects accelerate downwards at a rate of 9.8 m/s/s.
The Acceleration of Gravity
Topic sentence: The acceleration of gravity is shown as g and is numerically equal to 9.8 m/s/s.9.8 m/s/s is known as the acceleration of gravity and is symbolically shown as g. We will occasionally use the approximated value of 10 m/s/s in in order to reduce the complexity of the many mathematical tasks that we will perform with this number.
Representing Free Fall by Graphs
Topic sentence: Free falling objects have a negative acceleration.One mean of describing the motion of objects is through the use of graphs -position versus time and velocity vs. time graphs.
A curved line on a position versus time graph signifies an accelerated motion. A further look at the position-time graph reveals that the object starts slow and finishes fast. Finally, the negative slope of the line indicates a negative or downward velocity.
How Fast? and How Far?
Topic sentence: The velocity of a free falling object is dependent on the time it fell and the time of the fall.
The velocity of a free-falling object is changing by 9.8 m/s every second. The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is----vf = g * t
g is 9.8 m/s/s.
The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of t seconds is given by---d = 0.5 * g * t2
The Big Misconception
Topic sentence: A more massif object does not accelerate at a greater rate than a less massive object.
A more massive object does not accelerate at a greater rate than a less massive object. More massive objects will only fall faster if there is an appreciable amount of air resistance present. The actual explanation of why all objects accelerate at the same rate involves the concepts of force and mass. The greater force on more massive objects is offset by the inverse influence of greater mass, but all objects free fall at the same rate of acceleration, regardless of their mass.
10/4/11
Free Fall Lab
100 g weightThe equation of the velocity time graph of y=mx+b is the same as the formula v(f)=v(i)+at. m or the slope has the same value as a, or acceleration and b or the y intercept is equal to the initial velocity. Our slope or acceleration is 775.37 and inital velocity or y intercept is 52.763.
In the position time graph, the equation is y=Ax^2+Bx, is also equal to the equation d=v(i)t+.5at^2. In this equation A has the same value as half the acceleration and B has the same value as the initial velocity. In our equation, half our velocity or A value is 387.93 and initial velocity or B value is 52.006. The initial velocity should be zero but is not because the timer and the drop of the object might not have been at the same time, the object could have been in motion when the spark timer made the first dot.We used a polynomial shape trend line to also show that our results fit the trend line almost perfectly. Additionally, our r squared value is .99 which shows that our results were precise and correct.
Error is due to the friction that occurred with the spark timer.
Discussion Questions
1. Does the shape of your v-t graph agree with the expected graph? Why or why not?
The shape of our v-t graph fully agrees with the expected graph because they are both linear due to constant speed. Unfortunately, because the object is being dropped, the line should be moved negatively away from the origin. Because the negatives were left out of our data our graph turned out the way it did.
2. Does the shape of your x-t graph agree with the expected graph? Why or why not?
Yes, our x-t graph agrees with the expected graph, because the object accelerated downward and this is displayed in the curved graph.
3. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
Our final result was that the acceleration was 775.37 cm/s/s and the class average was 805.87 cm/s/s. After finding the percent difference, we found that we have a 3.78% difference. Our results were below the 5% difference range, which implies that they were very close to equal the class average.
4. Did the object accelerate uniformly? How do you know?
Yes, this is shown in our v-t graph because it is seen that it has a constant acceleration. This is because of the linear display that comes from our data. Although it did not accelerate at exactly 981 cm/s/s, it accelerated at a constant rate which was pretty close to this. This shows that it did accelerate uniformly.
5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
Additional force from the person dropping the weight could cause a higher acceleration due to gravity. However, in our case the lower acceleration of gravity was caused by the friction of the ticker tape with the spark timer as it was threaded through.
Conclusion:
Our final results were that our weight fell with an acceleration of 775.37 cm/s/s. However, the desired or theoretical result was 981 cm/s/s. We then found out that our percent error was 20.96% which implies accuracy, but unfortunately not precision. We also found that the percent difference from the class average was 3.78%, which is very close the class average proving that our results were almost accurate. Our hypothesis was also correct, however, we kept our acceleration due to gravity in meters rather than centimeters and we said that the result should be negative. This conclusion is correct because the object is falling down, but for the experiment we disregarded the negative. One main source of error in our experiment was the friction from the ticker tape going through the spark timer. This would cause the acceleration to be lower. To fix this for future trials, you could factor in the lower acceleration do to friction. There was also added error because human's are not machines, which causes an additional area of human error. Because I had to drop the mass off the balcony I could have been a source of error because I could have slightly pushed the mass downwards, which would change the initial velocity. Also, Rachel or I could have measured the distance between the dots on the ticker tape incorrectly. To fix this error we could have used a different form of measurement that is more precise.
Accelerated Motion